The equation of the normal to the curve y^2=x^3 at the point whose abscissa is 8 is?
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x = 8, so y2 = 8* 8* 8, so y = 8 root 8
slope of tangent = dy/dx, now 2y dy/dx = 3x2 or dy/dx = 3* 64 / 2* 8root8
= 12/ 2 root 2 = 3root 2, so slope of normal = -1 / 3root 2
now you can find the equation of the normal, using point and slope
y = mx + c, substitute the values to get c and the equation
slope of tangent = dy/dx, now 2y dy/dx = 3x2 or dy/dx = 3* 64 / 2* 8root8
= 12/ 2 root 2 = 3root 2, so slope of normal = -1 / 3root 2
now you can find the equation of the normal, using point and slope
y = mx + c, substitute the values to get c and the equation
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